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Question Number 281 by samarth last updated on 25/Jan/15
If f(x)=tan^(−1) (√((1+sin x)/(1−sin x))), 0≤x≤π/2, then f ′(π/6)=?
Iff(x)=tan−11+sinx1−sinx,0⩽x⩽π/2,thenf′(π/6)=?
Answered by 123456 last updated on 18/Dec/14
f(x)=tan^(−1) (√((1+sin x)/(1−sin x))) (∂f/∂x)=(∂/∂x)(tan^(−1) (√((1+sin x)/(1−sin x)))) =(1/(1+((√((1+sin x)/(1−sin x))))^2 ))∙(∂/∂x)((√((1+sin x)/(1−sin x)))) =(1/(1+((1+sin x)/(1−sin x))))∙(1/(2(√((1+sin x)/(1−sin x)))))∙(∂/∂x)(((1+sin x)/(1−sin x))) =(1/((1−sin x+1+sin x)/(1−sin x )))∙(1/2)(√(((1−sin x)/(1+sin x))∙))(((∂/∂x)(1+sin x)(1−sin x)−(1+sin x)(∂/∂x)(1−sin x))/((1−sin x)^2 )) =((1−sin x)/2)∙(1/2)(√(((1−sin x)/(1+sin x))∙))((cos x(1−sin x)+cos x(1+sin x))/((1−sin x)^2 )) =((1−sin x)/4)(√(((1−sin x)/(1+sin x))∙))((cos x−cos x sin x+cos x+cos x sin x)/((1−sin x)^2 )) =((1−sin x)/4)∙(((1−sin x)^(1/2) )/((1+sin x)^(1/2) ))∙((2cos x)/((1−sin x)^2 )) =(1/2)∙(((1−sin x)^(1/2) )/((1+sin x)^(1/2) ))∙((cos x)/(1−sin x)) x=(π/6)≡30° =(1/2)∙(((1−sin (π/6))^(1/2) )/((1+sin (π/6))^(1/2) ))∙((cos (π/6))/(1−sin (π/6))) =(1/2)∙(((1−(1/2))^(1/2) )/((1+(1/2))^(1/2) ))∙(((√3)/2)/(1−(1/2))) =(1/2)∙((((1/2))^(1/2) )/(((3/2))^(1/2) ))∙(((√3)/2)/(1/2)) =(1/2)∙((1/( (√2)))/((√3)/( (√2))))∙((√3)/2)∙2 =(1/2)∙(1/( (√2)))∙((√2)/( (√3)))∙(√3) =(1/2)
f(x)=tan−11+sinx1−sinx∂f∂x=∂∂x(tan−11+sinx1−sinx)=11+(1+sinx1−sinx)2⋅∂∂x(1+sinx1−sinx)=11+1+sinx1−sinx⋅121+sinx1−sinx⋅∂∂x(1+sinx1−sinx)=11−sinx+1+sinx1−sinx⋅121−sinx1+sinx⋅∂∂x(1+sinx)(1−sinx)−(1+sinx)∂∂x(1−sinx)(1−sinx)2=1−sinx2⋅121−sinx1+sinx⋅cosx(1−sinx)+cosx(1+sinx)(1−sinx)2=1−sinx41−sinx1+sinx⋅cosx−cosxsinx+cosx+cosxsinx(1−sinx)2=1−sinx4⋅(1−sinx)1/2(1+sinx)1/2⋅2cosx(1−sinx)2=12⋅(1−sinx)1/2(1+sinx)1/2⋅cosx1−sinxx=π6≡30°=12⋅(1−sinπ6)1/2(1+sinπ6)1/2⋅cosπ61−sinπ6=12⋅(1−12)1/2(1+12)1/2⋅321−12=12⋅(12)1/2(32)1/2⋅3212=12⋅1232⋅32⋅2=12⋅12⋅23⋅3=12

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